Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2]
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FROM(X) -> CONS(X, n_{_from}(n_{_s}(X)))
LENGTH(n_{_cons}(X, Y)) -> S(length1(activate(Y)))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))
LENGTH(n_{_cons}(X, Y)) -> ACTIVATE(Y)
LENGTH1(X) -> LENGTH(activate(X))
LENGTH1(X) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> FROM(activate(X))
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_s}(X)) -> S(activate(X))
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_nil}) -> NIL
ACTIVATE(n_{_cons}(X1, X2)) -> CONS(activate(X1), X2)
ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pairs:

ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_cons}(X1, X2)) -> ACTIVATE(X1)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__cons(x₁, x₂)) = 1 + x₁

POL(n__from(x₁)) = x₁

POL(n__s(x₁)) = x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pairs:

ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)
ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_s}(X)) -> ACTIVATE(X)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from(x₁)) = x₁

POL(n__s(x₁)) = 1 + x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Polo

...

               →DP Problem 4

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pair:

ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ACTIVATE(n_{_from}(X)) -> ACTIVATE(X)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from(x₁)) = 1 + x₁

POL(ACTIVATE(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Polo

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pairs:

LENGTH1(X) -> LENGTH(activate(X))
LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

LENGTH(n_{_cons}(X, Y)) -> LENGTH1(activate(Y))

five new Dependency Pairs are created:

LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(nil)
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rewriting Transformation

Dependency Pairs:

LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(nil)
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(X) -> LENGTH(activate(X))

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(nil)

one new Dependency Pair is created:

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(X) -> LENGTH(activate(X))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

LENGTH1(X) -> LENGTH(activate(X))

five new Dependency Pairs are created:

LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH1(n_{_nil}) -> LENGTH(nil)
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH1(X'') -> LENGTH(X'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 8

                 ↳Rewriting Transformation

Dependency Pairs:

LENGTH1(X'') -> LENGTH(X'')
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(n_{_nil}) -> LENGTH(nil)
LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

LENGTH1(n_{_nil}) -> LENGTH(nil)

one new Dependency Pair is created:

LENGTH1(n_{_nil}) -> LENGTH(n_{_nil})

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 9

                 ↳Forward Instantiation Transformation

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_nil})) -> LENGTH1(n_{_nil})
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(X'') -> LENGTH(X'')

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

LENGTH1(X'') -> LENGTH(X'')

five new Dependency Pairs are created:

LENGTH1(n_{_cons}(X'0, n_{_from}(X''''))) -> LENGTH(n_{_cons}(X'0, n_{_from}(X'''')))
LENGTH1(n_{_cons}(X'0, n_{_s}(X''''))) -> LENGTH(n_{_cons}(X'0, n_{_s}(X'''')))
LENGTH1(n_{_cons}(X''', n_{_cons}(X1''', X2'''))) -> LENGTH(n_{_cons}(X''', n_{_cons}(X1''', X2''')))
LENGTH1(n_{_cons}(X''', Y''')) -> LENGTH(n_{_cons}(X''', Y'''))
LENGTH1(n_{_cons}(X''', n_{_nil})) -> LENGTH(n_{_cons}(X''', n_{_nil}))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 10

                 ↳Forward Instantiation Transformation

Dependency Pairs:

LENGTH1(n_{_cons}(X''', n_{_nil})) -> LENGTH(n_{_cons}(X''', n_{_nil}))
LENGTH1(n_{_cons}(X''', Y''')) -> LENGTH(n_{_cons}(X''', Y'''))
LENGTH1(n_{_cons}(X''', n_{_cons}(X1''', X2'''))) -> LENGTH(n_{_cons}(X''', n_{_cons}(X1''', X2''')))
LENGTH1(n_{_cons}(X'0, n_{_s}(X''''))) -> LENGTH(n_{_cons}(X'0, n_{_s}(X'''')))
LENGTH1(n_{_cons}(X'0, n_{_from}(X''''))) -> LENGTH(n_{_cons}(X'0, n_{_from}(X'''')))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

LENGTH(n_{_cons}(X, Y')) -> LENGTH1(Y')

eight new Dependency Pairs are created:

LENGTH(n_{_cons}(X, n_{_from}(X''''))) -> LENGTH1(n_{_from}(X''''))
LENGTH(n_{_cons}(X, n_{_s}(X''''))) -> LENGTH1(n_{_s}(X''''))
LENGTH(n_{_cons}(X, n_{_cons}(X1''', X2'''))) -> LENGTH1(n_{_cons}(X1''', X2'''))
LENGTH(n_{_cons}(X, n_{_cons}(X'0'', n_{_from}(X'''''')))) -> LENGTH1(n_{_cons}(X'0'', n_{_from}(X'''''')))
LENGTH(n_{_cons}(X, n_{_cons}(X'0'', n_{_s}(X'''''')))) -> LENGTH1(n_{_cons}(X'0'', n_{_s}(X'''''')))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', n_{_cons}(X1''''', X2''''')))) -> LENGTH1(n_{_cons}(X''''', n_{_cons}(X1''''', X2''''')))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', Y'''''))) -> LENGTH1(n_{_cons}(X''''', Y'''''))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', n_{_nil}))) -> LENGTH1(n_{_cons}(X''''', n_{_nil}))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 11

                 ↳Polynomial Ordering

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_cons}(X''''', n_{_nil}))) -> LENGTH1(n_{_cons}(X''''', n_{_nil}))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', Y'''''))) -> LENGTH1(n_{_cons}(X''''', Y'''''))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', n_{_cons}(X1''''', X2''''')))) -> LENGTH1(n_{_cons}(X''''', n_{_cons}(X1''''', X2''''')))
LENGTH(n_{_cons}(X, n_{_cons}(X'0'', n_{_s}(X'''''')))) -> LENGTH1(n_{_cons}(X'0'', n_{_s}(X'''''')))
LENGTH(n_{_cons}(X, n_{_cons}(X'0'', n_{_from}(X'''''')))) -> LENGTH1(n_{_cons}(X'0'', n_{_from}(X'''''')))
LENGTH1(n_{_cons}(X''', n_{_cons}(X1''', X2'''))) -> LENGTH(n_{_cons}(X''', n_{_cons}(X1''', X2''')))
LENGTH1(n_{_cons}(X'0, n_{_s}(X''''))) -> LENGTH(n_{_cons}(X'0, n_{_s}(X'''')))
LENGTH1(n_{_cons}(X'0, n_{_from}(X''''))) -> LENGTH(n_{_cons}(X'0, n_{_from}(X'''')))
LENGTH(n_{_cons}(X, n_{_cons}(X1''', X2'''))) -> LENGTH1(n_{_cons}(X1''', X2'''))
LENGTH(n_{_cons}(X, n_{_s}(X''''))) -> LENGTH1(n_{_s}(X''''))
LENGTH(n_{_cons}(X, n_{_from}(X''''))) -> LENGTH1(n_{_from}(X''''))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(n_{_cons}(X''', Y''')) -> LENGTH(n_{_cons}(X''', Y'''))

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

The following dependency pairs can be strictly oriented:

LENGTH(n_{_cons}(X, n_{_s}(X''''))) -> LENGTH1(n_{_s}(X''''))
LENGTH(n_{_cons}(X, n_{_s}(X''))) -> LENGTH1(s(activate(X'')))

Additionally, the following usable rules for innermost can be oriented:

cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
s(X) -> n_{_s}(X)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
nil -> n_{_nil}

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__cons(x₁, x₂)) = 1

POL(n__from(x₁)) = 1

POL(from(x₁)) = 1

POL(activate(x₁)) = x₁

POL(cons(x₁, x₂)) = 1

POL(LENGTH1(x₁)) = x₁

POL(n__s(x₁)) = 0

POL(nil) = 0

POL(s(x₁)) = 0

POL(n__nil) = 0

POL(LENGTH(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_cons}(X''''', n_{_nil}))) -> LENGTH1(n_{_cons}(X''''', n_{_nil}))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', Y'''''))) -> LENGTH1(n_{_cons}(X''''', Y'''''))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', n_{_cons}(X1''''', X2''''')))) -> LENGTH1(n_{_cons}(X''''', n_{_cons}(X1''''', X2''''')))
LENGTH(n_{_cons}(X, n_{_cons}(X'0'', n_{_s}(X'''''')))) -> LENGTH1(n_{_cons}(X'0'', n_{_s}(X'''''')))
LENGTH(n_{_cons}(X, n_{_cons}(X'0'', n_{_from}(X'''''')))) -> LENGTH1(n_{_cons}(X'0'', n_{_from}(X'''''')))
LENGTH1(n_{_cons}(X''', n_{_cons}(X1''', X2'''))) -> LENGTH(n_{_cons}(X''', n_{_cons}(X1''', X2''')))
LENGTH1(n_{_cons}(X'0, n_{_s}(X''''))) -> LENGTH(n_{_cons}(X'0, n_{_s}(X'''')))
LENGTH1(n_{_cons}(X'0, n_{_from}(X''''))) -> LENGTH(n_{_cons}(X'0, n_{_from}(X'''')))
LENGTH(n_{_cons}(X, n_{_cons}(X1''', X2'''))) -> LENGTH1(n_{_cons}(X1''', X2'''))
LENGTH(n_{_cons}(X, n_{_from}(X''''))) -> LENGTH1(n_{_from}(X''''))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(n_{_cons}(X''', Y''')) -> LENGTH(n_{_cons}(X''', Y'''))

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 13

                 ↳Polynomial Ordering

Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_cons}(X''''', Y'''''))) -> LENGTH1(n_{_cons}(X''''', Y'''''))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', n_{_cons}(X1''''', X2''''')))) -> LENGTH1(n_{_cons}(X''''', n_{_cons}(X1''''', X2''''')))
LENGTH(n_{_cons}(X, n_{_cons}(X'0'', n_{_s}(X'''''')))) -> LENGTH1(n_{_cons}(X'0'', n_{_s}(X'''''')))
LENGTH1(n_{_cons}(X''', Y''')) -> LENGTH(n_{_cons}(X''', Y'''))
LENGTH(n_{_cons}(X, n_{_cons}(X'0'', n_{_from}(X'''''')))) -> LENGTH1(n_{_cons}(X'0'', n_{_from}(X'''''')))
LENGTH1(n_{_cons}(X''', n_{_cons}(X1''', X2'''))) -> LENGTH(n_{_cons}(X''', n_{_cons}(X1''', X2''')))
LENGTH1(n_{_cons}(X'0, n_{_from}(X''''))) -> LENGTH(n_{_cons}(X'0, n_{_from}(X'''')))
LENGTH(n_{_cons}(X, n_{_cons}(X1''', X2'''))) -> LENGTH1(n_{_cons}(X1''', X2'''))
LENGTH(n_{_cons}(X, n_{_from}(X''''))) -> LENGTH1(n_{_from}(X''''))
LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', n_{_nil}))) -> LENGTH1(n_{_cons}(X''''', n_{_nil}))

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

LENGTH1(n_{_s}(X'')) -> LENGTH(s(activate(X'')))

Additionally, the following usable rules for innermost can be oriented:

s(X) -> n_{_s}(X)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X
cons(X1, X2) -> n_{_cons}(X1, X2)
from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
nil -> n_{_nil}

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(n__from(x₁)) = 0

POL(from(x₁)) = 0

POL(n__cons(x₁, x₂)) = 0

POL(activate(x₁)) = x₁

POL(cons(x₁, x₂)) = 0

POL(LENGTH1(x₁)) = x₁

POL(n__s(x₁)) = 1

POL(nil) = 0

POL(s(x₁)) = 1

POL(n__nil) = 0

POL(LENGTH(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 14

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

LENGTH(n_{_cons}(X, n_{_cons}(X''''', Y'''''))) -> LENGTH1(n_{_cons}(X''''', Y'''''))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', n_{_cons}(X1''''', X2''''')))) -> LENGTH1(n_{_cons}(X''''', n_{_cons}(X1''''', X2''''')))
LENGTH(n_{_cons}(X, n_{_cons}(X'0'', n_{_s}(X'''''')))) -> LENGTH1(n_{_cons}(X'0'', n_{_s}(X'''''')))
LENGTH1(n_{_cons}(X''', Y''')) -> LENGTH(n_{_cons}(X''', Y'''))
LENGTH(n_{_cons}(X, n_{_cons}(X'0'', n_{_from}(X'''''')))) -> LENGTH1(n_{_cons}(X'0'', n_{_from}(X'''''')))
LENGTH1(n_{_cons}(X''', n_{_cons}(X1''', X2'''))) -> LENGTH(n_{_cons}(X''', n_{_cons}(X1''', X2''')))
LENGTH1(n_{_cons}(X'0, n_{_from}(X''''))) -> LENGTH(n_{_cons}(X'0, n_{_from}(X'''')))
LENGTH(n_{_cons}(X, n_{_cons}(X1''', X2'''))) -> LENGTH1(n_{_cons}(X1''', X2'''))
LENGTH(n_{_cons}(X, n_{_from}(X''''))) -> LENGTH1(n_{_from}(X''''))
LENGTH(n_{_cons}(X, n_{_cons}(X1', X2'))) -> LENGTH1(cons(activate(X1'), X2'))
LENGTH1(n_{_from}(X'')) -> LENGTH(from(activate(X'')))
LENGTH(n_{_cons}(X, n_{_from}(X''))) -> LENGTH1(from(activate(X'')))
LENGTH1(n_{_cons}(X1', X2')) -> LENGTH(cons(activate(X1'), X2'))
LENGTH(n_{_cons}(X, n_{_cons}(X''''', n_{_nil}))) -> LENGTH1(n_{_cons}(X''''', n_{_nil}))

Rules:

from(X) -> cons(X, n_{_from}(n_{_s}(X)))
from(X) -> n_{_from}(X)
length(n_{_nil}) -> 0
length(n_{_cons}(X, Y)) -> s(length1(activate(Y)))
length1(X) -> length(activate(X))
s(X) -> n_{_s}(X)
nil -> n_{_nil}
cons(X1, X2) -> n_{_cons}(X1, X2)
activate(n_{_from}(X)) -> from(activate(X))
activate(n_{_s}(X)) -> s(activate(X))
activate(n_{_nil}) -> nil
activate(n_{_cons}(X1, X2)) -> cons(activate(X1), X2)
activate(X) -> X

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:08 minutes

POL(n__cons(x₁, x₂))	= 1 + x₁
POL(n__from(x₁))	= x₁
POL(n__s(x₁))	= x₁
POL(ACTIVATE(x₁))	= x₁

POL(n__cons(x₁, x₂))	= 1
POL(n__from(x₁))	= 1
POL(from(x₁))	= 1
POL(activate(x₁))	= x₁
POL(cons(x₁, x₂))	= 1
POL(LENGTH1(x₁))	= x₁
POL(n__s(x₁))	= 0
POL(nil)	= 0
POL(s(x₁))	= 0
POL(n__nil)	= 0
POL(LENGTH(x₁))	= x₁